Chapter 7: Section 7-3 Permutations D. S. Malik Creighton University, Omaha, NE D. S. Malik Creighton University, Omaha, NE ()Chapter 7: Section 7-3 Permutations 1 / 13 Permutations Suppose that you are given four colored blocks-Red, Green, Blue, and Purple. How many ways can these four blocks, taken two at a time, be arranged in a row? We see the blocks and start making di¤erent arrangements and come up with the following arrangementsRG, RB, RP, GR, GB, GP, BR, BG, BP, PR, PG, and PB, which is a total of 12 di¤erent arrangements. These are called 2-permutations of the set of blocks. Notice that the arrangement RG is di¤erent than the arrangement GR. D. S. Malik Creighton University, Omaha, NE ()Chapter 7: Section 7-3 Permutations 2 / 13 Permutations If we are only interested in knowing the number of arrangements, then this activity can be completed in two steps. In the …rst step we select the …rst block and in the second step we select the second block. For the …rst step all four blocks are available and so the …rst step can be completed in 4 di¤erent ways. Because the …rst step has already used one of the blocks, for the second step only three blocks are available. Therefore, the second step can be completed in 3 di¤erent ways. By the multiplication principle the activity can be completed in 4 3 = 12 D. S. Malik Creighton University, Omaha, NE ()Chapter 7: Section 7-3 Permutations 3 / 13 De…nition Let S be a set with n distinct elements, n > 0. Let 0 < r n. An ordered arrangement of the elements of S taken r at a time is called an r -permutation of S. An n-permutation of S is called a permutation of S. De…nition Let S be a set with n distinct elements, n > 0. Let 0 < r P (n, r ) denotes the number of r -permutations of S. n. Then Theorem Let S be a set with n distinct elements, n > 0. Let 0 < r P (n, r ) is given by the following formula: P (n, r ) = n (n 1)(n 2) (n n. Then r + 1). Example Let us evaluate P (7, 3). Here n = 7 and r = 3. So n r + 1 = 7 3 + 1 = 5. Thus, P (7, 3) = 7 6 5 = 210. D. S. Malik Creighton University, Omaha, NE ()Chapter 7: Section 7-3 Permutations 4 / 13 Let n be a nonnegative integer. The factorial of n, written n!, is de…ned as follows: 0! = 1 n! = 1 2 (n 1) n = n (n 1) 2 1. For example, 3! = 1 2 3 = 6. Example (i) 5! = 1 2 3 4 5 = 120. (ii) (6 (iii) Let us evaluate 6! 6! 4)! = 2! = 1 2 = 2. 4!. 4! = (1 2 3 4 5 6) Form part (ii), it follows that (6 (1 2 3 4) = 720 4)! 6= 6! 24 = 696. 4!. D. S. Malik Creighton University, Omaha, NE ()Chapter 7: Section 7-3 Permutations 5 / 13 Theorem Let n and r be two integers such that 1 P (n, r ) = r n! (n r )! n. Then . Corollary Let S be a set with n distinct elements, n > 0. Then P (n, n )-the number of n-permutations of S is given by: P (n, n ) = n!. Example 13! 13! 13 12 11 10 9! = = (13 4)! 9! 9! = 13 12 11 10 = 17, 160. P (13, 4) = D. S. Malik Creighton University, Omaha, NE ()Chapter 7: Section 7-3 Permutations 6 / 13 Generalized Permutations Consider the word w = SUCCESS. w is a word consisting of the letters S, U, C , and E . The total number of letters in w is 7 and S occurs 3 times, C occurs 2 times, U occurs 1 time, and E occurs 1 time. Suppose we want to know all of the 7-letter distinct words consisting of letters S, U, C , and E such that S occurs 3 times, C occurs 2 times, U occurs 1 time, and E occurs one time. In other words, …nd the number of distinct permutations of the letters of the word SUCCESS. D. S. Malik Creighton University, Omaha, NE ()Chapter 7: Section 7-3 Permutations 7 / 13 Theorem Suppose there is a collection of n objects of k di¤erent types. Assume that objects of the same type cannot be distinguished from each other. Suppose that there are n1 objects of type 1, n2 objects of type 2, . . . , nk objects of type k, and n = n1 + n2 + + nk . Then the total number of di¤erent arrangements of these n objects of k di¤erent types taken all at a time is n1 !n2 ! n! nk 1 !nk ! n1 !n2 ! n! nk 1 !nk ! . De…nition Each of the arrangements of the above theorem is called a generalized arrangement or generalized permutation of the n objects of k di¤erent types taken all at a time. D. S. Malik Creighton University, Omaha, NE ()Chapter 7: Section 7-3 Permutations 8 / 13 Example Let us …nd the number of distinct permutations of the letters of the word SUCCESS. The total number of letters is n = 7. The distinct letters of this word are S, U, C , and E . The letter S occurs 3 times, C occurs 2 times, U occurs 1 time, and E occurs 1 time. Hence, the number of distinct permutations is 7! 7 6 5 4 3! 7! = = = 420. 3! 2! 1! 1! 3! 2! (3!) 2 1 D. S. Malik Creighton University, Omaha, NE ()Chapter 7: Section 7-3 Permutations 9 / 13 Exercise: How many dance pairs, (dance pairs means a pair (W , M ), where W stands for a woman and M for man), can be formed from a group of 6 women and 10 men? Solution: The problem is equivalent to …nding all 6-permutations of a set of 10 elements. Now the number of 6-permutations of a set of 10 elements is P (10, 6) = (1010!6 )! = 10! 4! = 10 9 8 7 6 5. D. S. Malik Creighton University, Omaha, NE ()Chapter 7: Section 7-3 Permutations 10 / 13 Exercise: In how many ways can 6 boys and 5 girls stand in a line so that no two girls are next to each other? Solution: According to the given conditions, between two girls there must be a boy. Suppose 6 boys are B1 , B2 , B3 , B4 , B5 , B6 and they stand in a line G B1 G B2 G B3 G B4 G B5 G B6 G where G ’s denote the positions for girls. For girls there are 7 positions. In these 7 positions, 5 di¤erent girls can stand in P (7, 5) di¤erent ways (because it is a 5-permutation of 7 elements). After each arrangement of the girls the 6 boys can stand in 6 di¤erent places in P (6, 6) di¤erent ways (because it is a 6-permutation of 6 elements). Hence, by the multiplication principle, the number of ways 6 boys and 5 girls may stand in a line so that no two girls are next to each other is 7! 7 6 5! 7! 6! = 6! = 6! = 7 6 P (7, 5) P (6, 6) = (7 2) ! 5! 5! D. S. Malik Creighton University, Omaha, NE ()Chapter 7: Section 7-3 Permutations 11 / 13 Exercise: Find the number of number of distinct permutations of the letters of the words ENGINEERING . Solution: In the word ENGINEERING , E occurs 3 times, N occurs 3 times, I occurs 2 times, G occurs 2 times and R occurs 1 time. Hence, the number of distinct permutations are 11! = 277, 200. 3! 3! 2! 2! 1! D. S. Malik Creighton University, Omaha, NE ()Chapter 7: Section 7-3 Permutations 12 / 13 Exercise: A sailing club has 10 white ‡ags, 7 red ‡ags, and 3 green ‡ags. In how many ways can these ‡ags be displayed in a row? Solution: The total number of ‡ags is 20. Out of these 20 ‡ags, 10 are white, 7 are red, and 3 are green. Hence, the number of order arrangements of these ‡ags in row is 20! = 22, 170, 720 10! 7! 3! D. S. Malik Creighton University, Omaha, NE ()Chapter 7: Section 7-3 Permutations 13 / 13